UNIT -5

APPLICATION OF PARTIAL DIFFERENTIAL EQUATIONS –

ONE DIMENSIONAL WAVE EQUATIONS .

APPLICATION OF PARTIAL DIFFERENTIAL EQUATION

 In physical problems, we always seek a solution of the differential
equation which satisfies some conditions known as boundary
conditions.
 The differential equation together with this boundary conditions,
contribute a boundary value problem.
PARTIAL DIFFERENTIAL EQUATIONS OF ENGINEERING
2 2
∂ y 2∂ y
1) Wave Equation: 2 =c 2
∂t ∂x
2
∂u 2 ∂ u
2) One Dimensional heat flow Equation: ∂ t =c 2
∂x
3) Two-Dimensional heat flow equation which in steady state becomes the
2 2
∂ u ∂ u
two-Dimensional Laplace’s Equation: 2
+ 2 =0
∂x ∂y

VIBRATIONS OF A STRETCHED STRING – WAVE EQUATION:

2 2
∂ y 2∂ y 2 T
2 where c= speed, T = tension ,
=c c=
∂t
2
∂x m

m= mass per unit length

y ( x , t ) why ? x = position , t- time

This the partial differentiation equation giving the transverse vibrations of the
string. It is also called the one-Dimensional wave equation.
SOLUTION OF THE WAVE EQUATION:
y=( C 1 ⅇ Px +C 2 ⅇ−Px )( C 3 ⅇC Pt + C4 e CPt )

y=( c 5 cos Px+ c 6 sin Px )( c 7 cos c P t +c 8 sin cPt )

y= ( C 9 x +C 10 )( C 11 t+ C12 )
 y= (C 1 cos P x +C 2 sin P x)(C 3 cos c P t+ C4 sin c P t) is the only suitable
solution of the wave equation.
 1-D Wave Equation – String
 1-D heat Equation – Rod
 2D heat Equation – Plate
BOUNDARY CONDITIONS:
As the end points of the string are fixed, for all time.
1) y(0,t)=0
2) y(l,t)=0
3) The initial transverse velocity of any point of the string is zero

( ∂∂ yt ) t =0
=0

4)y(x,0) = f(x) (Gn) Initial displacement.

THE GENERAL SOLUTION:
When velocity is given

( nπxl ) sin ( cnπtl )

∞

y(x,t)= ∑ bn sin
n =1

When velocity is not given

( nπxl ) cos ( cnπtl )

∞

y(x,t) =∑ bn sin
n =1

EXAMPLE : 1
A tightly stretched string with fixed end points x= 0 and x= l is the initially in
a position given by y = y 0 sin
3
( πxl ). If it is released from rest from this position,
find the displacement y(x,t).

Solution:
Step-1:
2 2
∂ y 2∂ y
The wave equation of the vibrating string is: 2 =c 2 --------(1)
∂t ∂x
 √
where c¿ T is the wave speed.
m

Step-2:
Boundary conditions, assume the string is fixed in both ends so,
y(0,t)=0
y(l,t)=0 for all t --------------(2)
This means the displacement must be zero at both ends at all times.
Step-3:
General solution with boundary conditions. The general solution
satisfying the boundary conditions is a Fourier sine series:

( nπxl )
∞

y(x,t) =∑ ¿¿ sin
n =1

Step-4:
Apply initial condition

 Initial displacement y= y 0 sin

3
( πxl ) --------------(3)
 Initial velocity
(velocity not given)

( ∂∂ yt ) t =0
=0 …………(4)

Since the vibration of the string is periodic therefore the solution of (1) is the
form of the
y =( c 1 cos Px+ c2 sin Px ) ( c 3 cos c Pt +c 4 sin cPt )

Here given x=0 , x=l

X=0 x=l
Substitute x=0
y(0,t) = ( c1 cos P(0)+c 2 sin P (0) ) ( c 3 cos cpt + c 4 sin cPt )
cos ( 0 )=1
 sin ( 0 )=0

y(0,t) = c 1 ( c 3 cos c P t +c 4 sin cPt )=0

c 1=0 this is to be true for all time
y(x,t) = (c ¿¿ 2 sin Px)( c3 cos cPt +c 4 sin cPt ) ¿
Also substitute x=l in the above equation
y(x,t) = (c ¿¿ 2 sin Px)( c3 cos cPt +c 4 sin cPt ) ¿

y(l,t) = (c ¿¿ 2 sin Pl) ( c 3 cos cpt +c 4 sin cPt ) ¿

nΠ
This gives pl=n π or p= l n being the integer

thus

y(x,t)= ¿ ¿sin ( nπxl ) ¿ ¿ cos ( cnπtl )+ ❑ sin ( cnπtl )…(5)

❑

Differentiate the above equation with respect to t,

∂y
∂t
=¿sin
nπx
l ( )
¿ ¿sin
cnπt
l
+ c 4 cos
l( )
cnπt
¿.
cnπ
l ( ) ( ) d¿

d(cosθ ¿=−sinθ
By (4)

( ∂∂ yt )
t =0
=¿ ¿sin ( nπxl ) ¿ ¿sin ( cnπl ( 0 ) )+ c cos ( cnπl ( 0 ) )¿ .( cnπl )
4

( ∂∂ yt )
t =0
=¿sin ( nπxl ) ¿ c ¿( cnπl )=0
4

Then c 4 =0
Substitute c 4 =0 Thus (5) becomes

y(x,t) = c2 sin l ( ( )) cos ( ))

nπx
¿
cnπt
l
 nπx
( ) ( )
cnπt
y(x,t) = c 2 c 3sin l cos l

So c 2 c 3=bn

y(x,t) = b nsin l( nπx ) cos( cnπtl ) (velocity not given)

Adding all such solutions of (1) is

( nπxl ) cos ( cnπtl )

∞

y(x,t) =∑ bn sin ……(6)

n =1

From (3)

( ) ( ) ( )
∞
πx nπx cnπt
y 0 sin3 =∑ b n sin cos
l n=1 l l

Put t=0

( ) ( ) cos ( 0 )
∞
πx nπx
y 0 sin3 =∑ b n sin
l n=1 l

sin3 ( )
πx
l
=¿
[ 3 sin ( πxl )−sin ( 3 πxl )
4 ]
( ) ( )
∞
πx nπx
y 0 sin3 =∑ b n sin
l n=1 l

Step5:
Expand the initial shape into a Fourier sine series

y0 [ 3 sin

4 ]
( πxl )−sin ( 3 πxl ) =b sin ( πxl ) + b sin ( 2 πxl )+ b sin ( 3 πxl ) +………..
1
2 3

Step 6:
Comparing both sides:
3 y0
b 1=
4
 b 2=0 ,

− y0
b 3= ,
4
b 4=0 ,

b 5=0

Final solution :

( ) ( )
3 y0 πx y0 3 πx
y(x,t) = 4 sin l + 0− 4 sin l

( ) ( ) ( ) ( )
3 y0 πx πct y0 3 πx 3 πct
y(x,t) = 4 sin l cos l − 4 sin l cos l

PRACTICE PROBLEMS:
A tightly stretched flexible string has its ends fixed at x=0 and x=l. at time t=o
string is given a shape defined by F(x)= μx ( l−x ) ,
Where μ is the constant, and then released. find the displacement of any point
X of the string at any time t¿ o .
Solution:
2 ∞
8μl 1 (2 m−1) π (2 m−1) πct
= 3
π
∑ (2 m−1)3 sin
l
x cos
l
m=1

EXAMPLE : 2
A tightly stretched string of length l with fixed ends is initially in

( )
3 πx
equilibrium position, it is set vibrating by giving each point a velocity v 0 sin l
. Find the displacement y ( x , t ) .
solution:
Step-1:
2 2
∂ y 2∂ y
The equation of the vibrating string is : 2 =c 2 ---------------------(1)
∂t ∂x

Step-2:
 Boundary conditions, assume the string is fixed in both ends so,
y(0,t)=0
y(l,t)=0 for all t --------------------------------
(2)
This means the displacement must be zero at both ends at all times,

and also the velocity is given as : ∂ t (∂ y) t =0

=v 0 sin3 ( πxl )

Step-3:
General solution with boundary conditions. The general solution
satisfying the boundary conditions is
y(x,t)=( c 1 cos px+ c 2 sin px )( c 3 cos cpt +c 4 sin cpt ) -----------------(3)
substitute y(0,t)
y(0,t)= ( c1 cos(0)+ c 2 sin(0) ) ( c3 cos cpt +c 4 sin cpt )
since sin 0 = 0,
cos 0 = 1
y(0,t)= c 1 ( c 3 cos cpt +c 4 sin cpt )
we know that y(0,t)=0 , then equating the above equation to zero. Therefore,
we get
c 1=0

Substitute c 1 in eq.(3)
y(x,t)= ( c 2 sin px ) ( c 3 cos cpt +c 4 sin cpt )--------------------------------(4)
Substitute y(l,t)
y(l,t)= ( c 2 sin pl ) ( c3 cos cpt +c 4 sin cpt )
we know that y(0,t)=0 , then equating the above equation to zero. Therefore,
we get

p= l( πx )
Substitute in eq.(4)

y(x,t)= c2 sin l( ( ))( ( )nπx

c 3 cos
cnπt
l
+c 4 sin
cnπt
l ( ))
0=c 2 c 3 sin ( nπxl ) cos ( cnπtl ) +c c sin ( nπxl ) sin( cnπtl )
2 4

Consider c 3 c 3=0, c 2 c 4=b n we get

nπx cnπt
y(x,t)= b n sin l sin l ( ) ( )
The general solution is

( nπxl ) sin ( cnπtl )

∞

y(x,t)= ∑ bn sin -------------------------------(5)

n =1

Step-4:
Differentiating the general solution with respect to ‘t’

( ) {∑ ( ) ( )( )}
∞
∂y nπx cnπt cnπ
= bn sin cos
∂t t =0 n=1 l l l t =0

( ) ( )
∞
πx cπ nπx
¿ ∑ n bn sin
3
v 0 sin
l l n=1 l

3 3 sin A−sin 3 A
Since sin A= 4

3 sin ( πxl )−sin ( 3 πxl )

sin
3
( πxl )= 4

v0 ( 3 sin

4
)
( πxl )−sin ( 3 πxl ) = cπ ∑ n b sin nπx
l (l)
∞

n=1
n

( ) ( ) ( ) ( ) ( )
3 v0 πx v 0 3 πx cπ πx 2 cπ 2 πx cπ 3 πx
sin − sin = b 1 sin + b 2 sin +3 b 3 sin
4 l 4 l l l l l l l

Step-5:
Comparing both sides and equating the same , we get
 ( ) ( )
3 v0 πx cπ πx
sin = b1 sin
4 l l l

3 v 0 cπ
= b1
4 l
3 v0l
b 1=
4 cπ
−v 0
4
sin ( )
3 πx
l
cπ
=3 b3 sin
l
3 πx
l ( )
−v 0 cπ
=3 b
4 l 3
−v 0 l
b 3= Whereas b 2 , b 4=0
12 cπ

Step-6:
Substituting the above values in the general solution (5)
We get,

( ) ( ) ( ) ( )
3 v0 l πx πct v0 l 3 πx 3 cπt
y(x,t)= 4 cπ sin l sin l − 12 cπ sin l sin l

y(x,t)= [ ( ) ( )
v0l 3
cπ 4
sin
πx
l
sin
πct
l
1
− sin
12
3 πx
l ( ) ( )]
sin
3 cπt
l

The final solution :

v 0l
[ ( ) ( ) ( ) ( )]
πx πct 3 πx 3 cπt
y(x,t)= 12cπ 9 sin l sin l −sin l sin l

PRACTICE PROBLEM:
A Tightly stretched string with fixed end points x=0, x=l initial at rest in its
equilibrium position. If it is vibrating by giving to each of its points with a
velocity λx(l-x). Find the displacement of the string at any distance x from the
end at any time ‘t’.
ANSWER:
∞
8 λ l3 1 (2 m−1) πx (2m−1) πct
4 ∑
y= 4
sin sin
c π m=1 (2 m−1) l l
 3
8 λl
Where b n= 4 4
c π ( 2 m−1 )